Difference between revisions of "2021 Fall AMC 12A Problems/Problem 2"

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~MRENTHUSIASM
 
~MRENTHUSIASM
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==Video Solution==
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https://youtu.be/8I2nzPmscTc
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~Charles3829
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==Video Solution (Simple and Quick)==
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https://youtu.be/Q3kkg-z0CXo
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~Education, the Study of Everything
  
 
==Video Solution==
 
==Video Solution==
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~savannahsolver
 
~savannahsolver
 
https://youtu.be/8I2nzPmscTc
 
 
~Charles3829
 
  
 
==Video Solution by TheBeautyofMath==
 
==Video Solution by TheBeautyofMath==
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~IceMatrix
 
~IceMatrix
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==Video Solution==
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https://youtu.be/aMkfa8VHjB8
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~Lucas
  
 
==See Also==
 
==See Also==

Latest revision as of 09:53, 18 September 2024

The following problem is from both the 2021 Fall AMC 10A #2 and 2021 Fall AMC 12A #2, so both problems redirect to this page.

Problem

Menkara has a $4 \times 6$ index card. If she shortens the length of one side of this card by $1$ inch, the card would have area $18$ square inches. What would the area of the card be in square inches if instead she shortens the length of the other side by $1$ inch?

$\textbf{(A) } 16 \qquad\textbf{(B) } 17 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 19 \qquad\textbf{(E) } 20$

Solution

We construct the following table: \[\begin{array}{c||c|c||c} & & & \\ [-2.5ex] \textbf{Scenario} & \textbf{Length} & \textbf{Width} & \textbf{Area} \\ [0.5ex] \hline & & & \\ [-2ex] \text{Initial} & 4 & 6 & 24 \\ \text{Menkara shortens one side.} & 3 & 6 & 18 \\ \text{Menkara shortens other side instead.} & 4 & 5 & 20 \end{array}\] Therefore, the answer is $\boxed{\textbf{(E) } 20}.$

~MRENTHUSIASM

Video Solution

https://youtu.be/8I2nzPmscTc

~Charles3829

Video Solution (Simple and Quick)

https://youtu.be/Q3kkg-z0CXo

~Education, the Study of Everything

Video Solution

https://youtu.be/158dTGn0zhI

~savannahsolver

Video Solution by TheBeautyofMath

for AMC 10: https://youtu.be/o98vGHAUYjM?t=100

for AMC 12: https://youtu.be/jY-17W6dA3c?t=101

~IceMatrix

Video Solution

https://youtu.be/aMkfa8VHjB8

~Lucas

See Also

2021 Fall AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
2021 Fall AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png